Hmmm... I can't see clearly what the continuation is without getting into heavy stuff (probably power series or something: I'm not sure), but it perhaps doesn't matter: the problem's title says geometry and, at least over here, this means the problem must be solved using exclusively tools from geometry, without any trigonometry, calculus, etc.
I wonder what the OP originally meant.

I first saw this problem in a paper of IMO preliminary selection contest but the suggested solution uses trigonometry . Of course , as i said it classic , there is actually a solution ( collected from a book )which only uses the properties of isosceles , equilateral triangle , similar triangles and parallel lines .

Below is the trigonometric method which is similar to the contest 's .

I first saw this problem in a paper of IMO preliminary selection contest but the suggested solution uses trigonometry . Of course , as i said it classic , there is actually a solution ( collected from a book )which only uses the properties of isosceles , equilateral triangle , similar triangles and parallel lines .

Below is the trigonometric method which is similar to the contest 's .

On the given triangle build a regular -gon such that is the center of it
(which is possible due to the fact that , ).
Now, we are not assuming . We'll come back to it later.
Continue AB until it gets to another vertex, call it , of the regular polygon (I'll leave it to you to verify the fact that it is indeed getting there), and draw a line as shown in the attached figure. Next, draw as shown, and call the intersection of with . All the angles shown in the figure can easily be calculated by considering equal arcs on the circumscribing circle of the regular polygon. Thus, from triangle it follows that , and in triangle we have . Therefore , but , so . Next notice that (again, by considering the circumscribing circle).
We have thus proven that the conditions and are obtained simultaneously, from which the result follows.

On the given triangle build a regular -gon such that is the center of it
(which is possible due to the fact that , ).
Now, we are not assuming . We'll come back to it later.
Continue AB until it gets to another vertex, call it , of the regular polygon (I'll leave it to you to verify the fact that it is indeed getting there), and draw a line as shown in the attached figure. Next, draw as shown, and call the intersection of with . All the angles shown in the figure can easily be calculated by considering equal arcs on the circumscribing circle of the regular polygon. Thus, from triangle it follows that , and in triangle we have . Therefore , but , so . Next notice that (again, by considering the circumscribing circle).
We have thus proven that the conditions and are obtained simultaneously, from which the result follows.

Wow , again ! A unbeatable method !

The key to this problem is : this isosceles triangle has the property that the difference of its base angle and the other angle is equal to 60 :

I would like to share my three solutions :

Spoiler:

Note that

Draw such that are on the same side of and .

Since , , ,

is congruent to and

.

Consider and , therefore , is an equilateral triangle .

Then , is an isosceles triangle with base angle , so .

Spoiler:

Let be a point outside the triangle such that is an equilateral triangle so . It is easy to find that is congruent to , . Note that is inside the . Now consider and , we have , and . Therefore , is congruent to . We can see that bisects so

Spoiler:

Let be a point inside the such that is an equilateral triangle , . It is easy to prove that is an isosceles trapezium . .